Get number of digits in an unsigned long integer c#

Question

I'm trying to determine the number of digits in ac# ulong number, i'm trying to do so using some math logic rather than using ToString().Length. I have not benchmarked the 2 approaches but have seen other posts about using System.Math.Floor(System.Math.Log10(number)) + 1 to determine the number of digits. Seems to work fine until i transition from 999999999999997 to 999999999999998 at which point, it i start getting an incorrect count.

Has anyone encountered this issue before ?

I have seen similar posts with a Java emphasis @ Why log(1000)/log(10) isn't the same as log10(1000)? and also a post @ How to get the separate digits of an int number? which indicates how i could possibly achieve the same using the % operator but with a lot more code

Here is the code i used to simulate this

Action<ulong> displayInfo = number => 
 Console.WriteLine("{0,-20} {1,-20} {2,-20} {3,-20} {4,-20}", 
  number, 
  number.ToString().Length, 
  System.Math.Log10(number), 
  System.Math.Floor(System.Math.Log10(number)),
  System.Math.Floor(System.Math.Log10(number)) + 1);

Array.ForEach(new ulong[] {
 9U,
 99U,
 999U,
 9999U,
 99999U,
 999999U,
 9999999U,
 99999999U,
 999999999U,
 9999999999U,
 99999999999U,
 999999999999U,
 9999999999999U,
 99999999999999U,
 999999999999999U,
 9999999999999999U,
 99999999999999999U,
 999999999999999999U,
 9999999999999999999U}, displayInfo);

Array.ForEach(new ulong[] {
 1U,
 19U,
 199U,
 1999U,
 19999U,
 199999U,
 1999999U,
 19999999U,
 199999999U,
 1999999999U,
 19999999999U,
 199999999999U,
 1999999999999U,
 19999999999999U,
 199999999999999U,
 1999999999999999U,
 19999999999999999U,
 199999999999999999U,
 1999999999999999999U
}, displayInfo);

Thanks in advance

Pat

Answer 1

log10 is going to involve floating point conversion - hence the rounding error. The error is pretty small for a double, but is a big deal for an exact integer!

Excluding the .ToString() method and a floating point method, then yes I think you are going to have to use an iterative method but I would use an integer divide rather than a modulo.

Integer divide by 10. Is the result>0? If so iterate around. If not, stop. The number of digits is the number of iterations required.

Eg. 5 -> 0; 1 iteration = 1 digit.

1234 -> 123 -> 12 -> 1 -> 0; 4 iterations = 4 digits.

Answer 2

I would use ToString().Length unless you know this is going to be called millions of times.

"premature optimization is the root of all evil" - Donald Knuth

Answer 3

From the documentation :

By default, a Double value contains 15 decimal digits of precision, although a maximum of 17 digits is maintained internally.

I suspect that you're running into precision limits. Your value of 999,999,999,999,998 probably is at the limit of precision. And since the ulong has to be converted to double before calling Math.Log10 , you see this error.

Answer 4

Other answers have posted why this happens.

Here is an example of a fairly quick way to determine the "length" of an integer (some cases excluded). This by itself is not very interesting -- but I include it here because using this method in conjunction with Log10 can get the accuracy "perfect" for the entire range of an unsigned long without requiring a second log invocation.

// the lookup would only be generated once
// and could be a hard-coded array literal
ulong[] lookup = Enumerable.Range(0, 20)
    .Select((n) => (ulong)Math.Pow(10, n)).ToArray();
ulong x = 999;
int i = 0;
for (; i < lookup.Length; i++) {
    if (lookup[i] > x) {
        break;
    }
}
// i is length of x "in a base-10 string"
// does not work with "0" or negative numbers

This lookup-table approach can be easily converted to any base. This method should be faster than the iterative divide-by-base approach but profiling is left as an exercise to the reader. (A direct if-then branch broken into "groups" is likely quicker yet, but that's way too much repetitive typing for my tastes.)

Happy coding.